Let $U$ be a $G_{\delta}$-set. (That is, $U$ is a countable intersections of open sets) such that it has following properties:
$U \subset [a, b]$,
$\mathbb Q\cap [a,b] \subset U$,
$m(U)>0$, where $m$ is the Lebesgue measure on $[a,b]$.
My question is: Will the translation of $U$ by rationals cover $\Bbb{R}$, i.e. is $$ \bigcup_{q \in \mathbb{Q}}(q+ U)=\Bbb{R} $$ true? $U$ must be uncountable as it has positive measure. But it may has empty interior. I need help on this question.
No, your assertion is not true in general. To give a counterexample, let $K$ be the set of algebraic numbers and $D = (K \setminus \mathbb Q) \cap (a, b)$. Note that $D$ is countable infinite. Let
$$U = \bigcap_{d\in D} (a, b)\setminus \{d\}.$$
This $U$ is a $G_\delta$ set in $[a, b]$ which contains the rationals in $(a, b)$ with positive measure. But
$$\bigcup_{q\in \mathbb Q} (q+ U)$$
and is not the whole $\mathbb R$: for example it does not contain $\sqrt 2$ (Reason: If $\sqrt 2 \in \bigcup_{q\in \mathbb Q} (q+ U)$, then
$$(*)\ \ \ \ \sqrt 2 = p + t$$
for some $p\in \mathbb Q$ and $t\in U$. But $t\in U$ imply that either $t$ is rational or $t$ is transcendental. The first case would give that $\sqrt 2$ is rational (which is absurd), while the second case is also impossible as $\sqrt 2 - p$ is algebraic)